this part of finding the limit [in this form] ALWAYS confuses me, i really can get it [maybe the (h) throws me off] the limit of [f(x+h)-f(x)] / h 1. f(x) = 3x-1 How does 3x-1 substitute in when there is an "h", it is confusing
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Hello, Originally Posted by >_<SHY_GUY>_< this part of finding the limit [in this form] ALWAYS confuses me, i really can get it [maybe the (h) throws me off] the limit of [f(x+h)-f(x)] / h 1. f(x) = 3x-1 How does 3x-1 substitute in when there is an "h", it is confusing $\displaystyle f({\color{red}x})=3{\color{red}x}-1$ $\displaystyle f({\color{red}x+h})=3({\color{red}x+h})-1$ So $\displaystyle \frac{f(x+h)-f(x)}{h}=\frac{[3(x+h)-1]-[3x-1]}{h}=\frac{3x+3h-1-3x+1}{h}=\frac{3h}{h}$ Simplify
$\displaystyle \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{[3(x+h)-1]-(3x-1)}{h}=\lim_{h\to 0}\frac{3x+3h-1-3x+1}{h}=$ $\displaystyle =\lim_{h\to 0}\frac{3h}{h}=3$
Originally Posted by Moo Hello, $\displaystyle f({\color{red}x})=3{\color{red}x}-1$ $\displaystyle f({\color{red}x+h})=3({\color{red}x+h})-1$ So $\displaystyle \frac{f(x+h)-f(x)}{h}=\frac{[3(x+h)-1]-[3x-1]}{h}=\frac{3x+3h-1-3x+1}{h}=\frac{3h}{h}$ Simplify hmm...right...but it says [and i forgot to mention that x approaches zero [the limit] the answer 3?
nevermind....errrrr im so dumb...thank you, i get it now haha
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